Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in JAVA):

Solution:
// upsolve with rainboy
import java.io.*;
import java.util.*;

public class CF1016E {
	static int[] ll, rr, aa;
	static double solve(int n, long u, long v) {
		int i, lower = -1, upper = n;
		while (upper - lower > 1) {
			i = (lower + upper) / 2;
			if (ll[i] * v <= u)
				lower = i;
			else
				upper = i;
		}
		i = lower;
		if (i == -1)
			return 0;
		return (i == 0 ? 0 : aa[i - 1]) + (double) Math.min(u, rr[i] * v) / v - ll[i];
	}
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter pw = new PrintWriter(System.out);
		StringTokenizer st = new StringTokenizer(br.readLine());
		int ys = -Integer.parseInt(st.nextToken());
		int xa = Integer.parseInt(st.nextToken());
		int xb = Integer.parseInt(st.nextToken());
		int n = Integer.parseInt(br.readLine());
		ll = new int[n];
		rr = new int[n];
		aa = new int[n];
		for (int i = 0; i < n; i++) {
			st = new StringTokenizer(br.readLine());
			ll[i] = Integer.parseInt(st.nextToken());
			rr[i] = Integer.parseInt(st.nextToken());
			aa[i] = rr[i] - ll[i] + (i == 0 ? 0 : aa[i - 1]);
		}
		int q = Integer.parseInt(br.readLine());
		while (q-- > 0) {
			st = new StringTokenizer(br.readLine());
			long x = Long.parseLong(st.nextToken());
			long y = Long.parseLong(st.nextToken());
			long ua = x * ys + xa * y;
			long ub = x * ys + xb * y;
			long v = ys + y;
			pw.println((solve(n, ub, v) - solve(n, ua, v)) * v / y);
		}
		pw.close();
	}
}